#### Knowledge Check 1: Simulation Perspective

This simulation demonstrates the relationships between Right Ascension, Sidereal Time, and Hour Angle. You should think of this simulation as relating ... where an object is on the celestial sphere (Right Anscension), to where it would be seen in our sky (Hour Angle), at a particular time (Sidereal Time). Information will be presented on each of these quantities separately before they are used in combination.

Spend a minute familiarizing yourself with the view provided in the simulation. In which direction are you looking? Describe how the observer's meridian appears in this simulation. What is the semicircle connecting the east point on the horizon to the west point of the horizon called? Note that the east point on the horizon is found in the far left of the simulation and the west point is on the far right. Where is the zenith in this simulation? You can think of this simulation as everything you can see from your peripheral vision furthest to the right to your peripheral vision furthest to the left when you are looking at the southern horizon -- all unfurled onto a flat surface.

There are a number of simplifications/assumptions used in the simulation.

- It is assumed that the observer is a mid-northern hemisphere observer looking at the south point of the horizon.
- Is is assumed that all objects are on the celestial equator. In other words we will ignore the declination of all objects and just focus on their right ascension (basically considering the declination to be zero).

#### Knowledge Check 2: Right Ascension

Right ascension describes where objects are located on the celestial sphere. It is analogous to longitude on Earth, but RA is measured in time units and only increases in one direction (what astronomers call east). Right ascension ranges from 0^{h}0^{m}0^{s} to 23^{h}59^{m}59^{s}.

The active star initially has an RA of 6^{h}. Use the RA dial to vary the RA of the active star. Move the star to an RA of 22 hours.

There are some important locations on the celestial sphere describing important locations for the sun that we will make use of in this module:

- Vernal Equinox --~March 21 --the sun has an RA of 0 hours and is on the celestial sphere.
- Summer Solstice -- ~June 21 -- the sun has an RA of 6 hours and is north (above) of the celestial equator.
- Autumnal Equinox -- ~September 21 --the sun has an RA of 12 hours and is on the celestial sphere.
- Winter Solstice -- ~December 21 -- the sun has an RA of 18 hours and is south (below) of the celestial equator.

#### Knowledge Check 3: Hour Angle

Hour Angle describes an objects relationship with the meridian of the observer in terms of time. All objects seen in this simulation will rise in the east, cross the observer's meridian when halfway across the sky, and set in the west. Hour angle describes where an object is in this process. When an object crosses the observer's meridian it has an hour angle of zero. Hour angle is measured in hours. When an object is east of the meridian, it has a negative hour angle -- the amount of time until the object crosses the observer's meridian due to Earth's rotation.

Move the active star to a RA of 2 hours. Now move the HA to -2 hours. How long until this object will cross the meridian. How long ago did the star rise? What is the HA of the star as it sets?

#### Knowledge Check 4: Sidereal Time

Sidereal time is time with respect to the stars. Realize that a typical 24 hour "solar" day depends on Earth's rotation and a little on its revolution. A sidereal day depends only upon rotation. So picture Earth isolated in space and spinning at is present rate -- it would take 23 hours and 56 minutes for one complete rotation.

Local sidereal time is defined as the RA value presently on the meridian. Start with the star on the eastern horizon. Simulate the normal apparent motion of the star in the sky -- from rising to setting --by changing the LST.

Change the local sidereal time to 4 hours. What is the RA of the CE location on the meridan?

The three quantaties are related in the following equation: **Hour Angle = Local Sidereal Time - Right Ascension**

#### Knowledge Check 5:Worked Story Problems

**Problem 1**: If a star has an RA of 4 hours and Earth's rotation carried it past the meridian 2 hours ago, what is the LST?

Solution: Note that the HA of the star is 2 hours (it is 2 hours past -- west -- of the meridian). So an RA of 4 hours is 2 hours west of the meridian and the RA on the meridian must be 6 hours since RA increases to the east. Check this answer in the simulator.

**Problem 2**: If it is midnight on the summer solstice, what is the hour angle of a star on the celestial equator with an RA of 16 hours?

Solution: If it is the summer solstice, the sun has an RA of 6 hours. If it is midnight, the sun is in the opposite direction as the meridian, so the LST is 18 hours. So move the active star to an RA of 16 hours, and read off the HA -- which is 2 hours. This star passed the meridian 2 hours ago and sets in 4 hours. Check this answer in the simulator.

#### Knowledge Check 6: More Story Problems

**Problem 3**:If a star has an RA of 9 hours and in its motion across the sky, 3 hours have passed since it crossed the observer's meridian. What is the local sidereal time?

**Problem 4**:If the local sidereal time is 7 hours, what is the HA of a star with an RA of 10 hours?

**Problem 5**:If it is 6 pm on the autumnal equinox, what is the RA of a star on the celestial equator with an HA of 3 hours?

**Problem 6**:If the vernal equinox has an hour angle of 2 hours, what is the local sidereal time?